Question

Steam at 4 MPa, 400°C enters a steady-flow, adiabatic turbine through a 20 cm-diameter-pipe with a velocity of 20 m/s. It leaves this turbine at 50 kPa with a quality of 80% through a 1 m-diameter pipe. What is the velocity of the steam as it leaves the turbine? (a) 10.3 m/s (b) 28.2 m/s (c) 32.6 m/s (d) 73.3 m/s

Answer 1

Answer:

28,8 m/s

Explanation:

In a steady flow system we can say that m1=m2 which means that the mass flow in the entrance in the same in the outlet.  m is flow (kg/s)

we know that $m=\frac{1}{v} A*V$ where V (m/s) is velocity, A (m^2) ia area and v is specific volume (m^3/kg)

Since m1=m2 we can say

$\frac{1}{v_{1} } A_{1} *V_{1}=\frac{1}{v_{2}} A_{2} *V_{2}$

clearing the equation

$V_{2}= \frac{v_{2} }{v_{1}}*\frac{A_{1} }{A_{2}} *V_{1}$

we can specific volume (m^3/kg) from thermodynamic tables

for the entrance is 400°C and 4 MPa is superheated steam and v is : 0,7343 m^3/kg

In the outlet we have saturated vapor with quality (x) of 80%. In this case we get the specific saturated volume for the liquid (vf) and the specific volume for the saturated  (vg) gas from the thermodynamic tables. we use the next equation to get  (v) for the condition of interest, in this case 80% quality.

v= vf +x*(vg - vf)

where:

x: quality

vf = liquid-saturated-specific-volume

vg =steam-saturated-specific-volume.

for this problem

x = 0,8

vf = 0,00102991

vg = 3,24015

so

we get = 2,593 m^3/kg

The area is the one for a circle

$\pi *r^{2}$

r1 = 0,1 m^2 for area 1

r2=0,5 m^2 for area 2

A1 = 0,0314 m^2

A2 = 0,7853 m^2

we know that  V1 is 20 m/s

replacing these values in the equation

$V_{2}= \frac{v_{2} }{v_{1}}*\frac{A_{1} }{A_{2}} *V_{1}$

we get V2 = 28,2 m/s.